Figure It Out 1.2

Question 1

Find the cube roots of 27000 and 10648.

Solution:

1. 27000: $27000 = 27 \times 1000 = 3^3 \times 10^3$. $\sqrt[3]{27000} = 3 \times 10 = 30$.

2. 10648: Prime factorization: $10648 \div 8 = 1331$ $1331 \div 11 = 121$ $121 \div 11 = 11$ Factors: $(2 \times 2 \times 2) \times (11 \times 11 \times 11)$. $\sqrt[3]{10648} = 2 \times 11 = 22$.

Question 2

What number will you multiply by 1323 to make it a cube number?

Solution: Prime factorization of 1323: $1323 \div 3 = 441$ $441 \div 3 = 147$ $147 \div 3 = 49$ $49 = 7 \times 7$ Factors: $(3 \times 3 \times 3) \times 7 \times 7$. The 3s form a triplet. The 7s are only two ($7 \times 7$). We need one more 7 to complete the triplet.

Answer: Multiply by 7.

Question 3

State true or false. (i) The cube of any odd number is even. False. $3^3 = 27$ (odd). (ii) There is no perfect cube that ends with 8. False. $2^3 = 8$, $12^3 = 1728$. (iii) The cube of a 2-digit number may be a 3-digit number. False. Smallest 2-digit number is 10. $10^3 = 1000$ (4 digits). (iv) The cube of a 2-digit number may have seven or more digits. False. Largest 2-digit number is 99. $99 < 100$. $100^3 = 1,000,000$ (7 digits). So $99^3$ must have fewer than 7 digits (it has 6). (v) Cube numbers have an odd number of factors. False. Only squares have an odd number of factors. E.g., $8$ has factors 1, 2, 4, 8 (4 factors). Exception: If a number is both a square and a cube (like 64), it has odd factors. But generally, no.

Question 4

Guess the cube roots of 4913, 12167, and 32768.

Solution:

• 4913: Ends in 3 $\to$ Root ends in 7. Strike out last 3 digits (913). Remaining is 4. Largest cube $<4$ is $1^3=1$. Root: 17.
• 12167: Ends in 7 $\to$ Root ends in 3. Strike out 167. Remaining 12. Largest cube $<12$ is $2^3=8$. Root: 23.
• 32768: Ends in 8 $\to$ Root ends in 2. Strike out 768. Remaining 32. Largest cube $<32$ is $3^3=27$. Root: 32.

Question 5

Which is the greatest? $67^3 - 66^3$, etc.

Solution: We can use the difference formula approximation or calculate pattern: $n^3 - (n-1)^3 = 1 + n(n-1) \times 3$.

• (i) $67^3 - 66^3 \approx 1 + 67 \times 66 \times 3 \approx 13,000$
• (ii) $43^3 - 42^3$ is smaller (smaller $n$).
• (iii) $67^2 - 66^2 = 67 + 66 = 133$. (Much smaller).
• (iv) $43^2 - 42^2 = 85$.

Answer: (i) $67^3 - 66^3$ is the greatest.